Given a positive integer n. The problem is to print numbers in the range 1 to n having first and last bits as the only set bits.
Examples:

Input : n = 10
Output : 1 3 5 9
(1)₁₀ = (1)₂.
(3)₁₀ = (11)₂.
(5)₁₀ = (101)₂.
(9)₁₀ = (1001)₂

Naive Approach: Print "1". Now for i = 3 to n, check if (i-1) is a Perfect power of two or not. If true then print i.

C++

// C++ implementation to print numbers in the range 1 to n
// having first and last bits as the only set bits
#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long int ull;

// function to check whether 'n'
// is a power of 2 or not
bool powerOfTwo(ull n)
{
    return (!(n & n-1));
}

// function to print numbers in the range 1 to n having
// first and last bits as the only set bits
void printNumWithFirstLastBitsSet(ull n)
{
    ull i = 1;
    
    // first number is '1'    
    cout << i << " ";
    
    // generating all the numbers
    for (i = 3; i <= n; i++)
        // if true, then print 'i'
        if (powerOfTwo(i-1))
            cout << i << " ";    
}

// Driver program to test above
int main()
{
    ull n = 10;
    printNumWithFirstLastBitsSet(n);
    return 0;
}

Java

// Naive approach 
// Java implementation to print
// numbers in the range 1 to n
// having first and last bits as
// the only set bits
import java.io.*;

class GFG {
    
    // function to check whether 'n'
    // is a power of 2 or not
    static Boolean powerOfTwo(long n)
    {
        return (!((n & n-1) != 0));
    }
    
    // function to print numbers in the 
    // range 1 to n having first and
    // last bits as the only set bits
    static void printNumWithFirstLastBitsSet(long n)
    {
        long i = 1;
        
        // first number is '1' 
        System.out.print( i + " ");
        
        // generating all the numbers
        for (i = 3; i <= n; i++)
        
            // if true, then print 'i'
            if (powerOfTwo(i - 1))
                System.out.print(i + " "); 
    }
        // Driver function
        public static void main (String[] args) {
        long n = 10l;
        printNumWithFirstLastBitsSet(n);
    
        }
}

//This code is contributed by Gitanjali.

Python3

# Python implementation to print 
# numbers in the range 1 to n
# having first and last bits 
# as the only set bits
import math

# function to check whether 'n'
# is a power of 2 or not
def powerOfTwo(n):
    re = (n & n - 1)
    return (re == 0)

# function to print numbers 
# in the range 1 to n having
# first and last bits as 
# the only set bits
def printNumWithFirstLastBitsSet(n):
    i = 1
    
    # first number is '1' 
    print ( i, end = " ")
    
    # generating all the numbers
    for i in range(3, n + 1):
        
        # if true, then print 'i'
        if (powerOfTwo(i - 1)):
            print ( i, end = " ")
    
    
# driver function
n = 10
printNumWithFirstLastBitsSet(n)

# This code is contributed by Gitanjali.

// Naive approach 
// C# implementation to print
// numbers in the range 1 to n
// having first and last bits as
// the only set bits
using System;

class GFG {
    
    // function to check whether 'n'
    // is a power of 2 or not
    static Boolean powerOfTwo(long n)
    {
        return (!((n & n-1) != 0));
    }
    
    // function to print numbers in the 
    // range 1 to n having first and
    // last bits as the only set bits
    static void printNumWithFirstLastBitsSet(long n)
    {
        long i = 1;
        
        // first number is '1' 
        Console.Write( i + " ");
        
        // generating all the numbers
        for (i = 3; i <= n; i++)
        
            // if true, then print 'i'
            if (powerOfTwo(i - 1))
                Console.Write(i + " "); 
    }
    
    // Driver function
    public static void Main () 
    {
        long n = 10L;
        printNumWithFirstLastBitsSet(n);
        
    }
}

// This code is contributed by Vt_m.

PHP

<?php
// php implementation to print 
// numbers in the range 1 to n
// having first and last bits 
// as the only set bits

// function to check whether 'n'
// is a power of 2 or not
function powerOfTwo($n)
{
    return (!($n & $n - 1));
}

// function to print numbers in
// the range 1 to n having
// first and last bits as
// the only set bits
function printNumWithFirstLastBitsSet($n)
{
    $i = 1;
    
    // first number is '1' 
    echo $i." ";
    
    // generating all the numbers
    for ($i = 3; $i <= $n; $i++)
    
        // if true, then print 'i'
        if (powerOfTwo($i - 1))
            echo $i." "; 
}

    // Driver Code
    $n = 10;
    printNumWithFirstLastBitsSet($n);

// This Code is contributed by mits 
?>

JavaScript

<script>

// Javascript implementation to print numbers in the range 1 to n
// having first and last bits as the only set bits

// function to check whether 'n'
// is a power of 2 or not
function powerOfTwo(n)
{
    return (!(n & n-1));
}

// function to print numbers in the range 1 to n having
// first and last bits as the only set bits
function printNumWithFirstLastBitsSet(n)
{
    let i = 1;
    
    // first number is '1'
    document.write(i + " ");
    
    // generating all the numbers
    for (i = 3; i <= n; i++)
        // if true, then print 'i'
        if (powerOfTwo(i-1))
            document.write(i + " ");
}

// Driver program to test above

    let n = 10;
    printNumWithFirstLastBitsSet(n);


//This code is contributed by Mayank Tyagi
</script>

Output:

1 3 5 9

Time Complexity : O(n)

Auxiliary Space: O(1)

Efficient Approach: Print "1". Now one by one generate perfect power of two (except '1') with the help of bitwise left shift operation. Bitwise xor these numbers with 1 and if result is in the range print them else stop.

C++

// C++ implementation to print numbers in the range 1 to n
// having first and last bits as the only set bits
#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long int ull;

// function to print numbers in the range 1 to n having
// first and last bits as the only set bits
void printNumWithFirstLastBitsSet(ull n)
{
    ull power_2 = 1, num;
    
    // first number is '1'    
    cout << power_2 << " ";
    
    while (1)    
    {
        // obtaining next perfect power of 2
        power_2 <<= 1;
        
        // toggling the last bit to convert 
        // it to as set bit
        num = power_2 ^ 1;
        
        // if out of range then break;
        if (n < num)
            break;
            
        // display    
        cout << num << " ";    
    }        
}

// Driver program to test above
int main()
{
    ull n = 10;
    printNumWithFirstLastBitsSet(n);
    return 0;
}

Java

// efficient approach Java implementation
// to print numbers in the range 1 to n
// having first and last bits as the only set bits

import java.io.*;

class GFG {

    // function to print numbers in 
    // the range 1 to n having first and 
    // last bits as the only set bits
    static void prNumWithFirstLastBitsSet(long n)
    {
        long power_2 = 1, num;
    
        // first number is '1' 
        System.out.print(power_2 + " ");
    
        while (true) 
        {
            // obtaining next perfect power of 2
            power_2 <<= 1;
            
            // toggling the last bit to  
            // convert it to as set bit
            num = power_2 ^ 1;
            
            // if out of range then break;
            if (n < num)
                break;
                
            // display 
            System.out.print(num + " "); 
        }
    
}

    public static void main (String[] args) {
    long n = 10;
    prNumWithFirstLastBitsSet(n);

    }
}
// This code is contributed by Gitanjali.

Python3

# Python3 implementation to 
# pr numbers in the range 
# 1 to n having first and 
# last bits as the only set bits


# function to print numbers in the 
# range 1 to n having first and 
# last bits as the only set bits
def prNumWithFirstLastBitsSet(n):
    
    power_2 = 1
    
    # first number is '1' 
    print ( power_2, end = ' ')
    
    while (1):
        # obtaining next perfect
        # power of 2
        power_2 <<= 1
        
        # toggling the last bit to  
        # convert it to as set bit
        num = power_2 ^ 1
        
        # if out of range then break;
        if (n < num):
            break
            
        # display 
        print ( num, end = ' ') 
    


# Driver program 
n = 10;
prNumWithFirstLastBitsSet(n)

# This code is contributed by saloni1297

// efficient approach C# implementation
// to print numbers in the range 1 to n
// having first and last bits as the only set bits
using System;

class GFG {

    // function to print numbers in 
    // the range 1 to n having first and 
    // last bits as the only set bits
    static void prNumWithFirstLastBitsSet(long n)
    {
        long power_2 = 1, num;
    
        // first number is '1' 
        Console.Write(power_2 + " ");
    
        while (true) 
        {
            // obtaining next perfect power of 2
            power_2 <<= 1;
            
            // toggling the last bit to 
            // convert it to as set bit
            num = power_2 ^ 1;
            
            // if out of range then break;
            if (n < num)
                break;
                
            // display 
            Console.Write(num + " "); 
        }
    
    }
    
    // Driver code
    public static void Main () 
    {
        long n = 10;
        prNumWithFirstLastBitsSet(n);

    }
}
// This code is contributed by vt_m.

PHP

<?php
// php implementation to print 
// numbers in the range 1 to n
// having first and last bits 
// as the only set bits

// function to print numbers in 
// the range 1 to n having
// first and last bits as 
// the only set bits
function printNumWithFirstLastBitsSet($n)
{
    $power_2 = 1;
    
    // first number is '1' 
    echo $power_2." ";
    
    while (1) 
    {
        
        // obtaining next perfect
        // power of 2
        $power_2 <<= 1;
        
        // toggling the last
        // bit to convert 
        // it to as set bit
        $num = $power_2 ^ 1;
        
        // if out of range 
        // then break;
        if ($n < $num)
            break;
            
        // display 
        echo $num." "; 
    } 
}

    // Driver code
    $n = 10;
    printNumWithFirstLastBitsSet($n);

// This code is contributed by mits 
?>

JavaScript

<script>

// Javascript implementation to print numbers in the range 1 to n 
// having first and last bits as the only set bits 

// function to print numbers in the range 1 to n having 
// first and last bits as the only set bits 
function printNumWithFirstLastBitsSet(n) 
{ 
    var power_2 = 1, num; 
    
    // first number is '1'     
    document.write(power_2 + " "); 
    
    while (true)     
    { 
        // obtaining next perfect power of 2 
        power_2 <<= 1; 
        
        // toggling the last bit to convert 
        // it to as set bit 
        num = power_2 ^ 1; 
        
        // if out of range then break; 
        if (n < num) 
            break; 
            
        // display     
        document.write(num + " ");     
    }         
} 

// Driver program to test above 
var n = 10; 
printNumWithFirstLastBitsSet(n); 
    
</script>

Output:

1 3 5 9

Time Complexity : O(logn)

Auxiliary Space: O(1)

Print numbers having first and last bits as the only set bits

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